## Algebra 2 (1st Edition)

$\approx0.4$ seconds
The ball lands on the opposite side when $h=0$. So using the given formula with $h_0=9$ and $v_0=-16$ we get: $0=-16t^2-16t+9$ Then using the quadratic formula: $t=\frac{16\pm\sqrt{(-16)^2-4(-16)9}}{2(-16)}=\frac{-2\pm\sqrt{13}}{4}$ We consider only the positive answer: $t=\frac{2+\sqrt{13}}{4}\approx0.4$